3.2443 \(\int \frac {1}{(a+\frac {b}{\sqrt [3]{x}})^3 x^4} \, dx\)

Optimal. Leaf size=146 \[ -\frac {84 a^6 \log \left (a \sqrt [3]{x}+b\right )}{b^9}+\frac {28 a^6 \log (x)}{b^9}+\frac {21 a^6}{b^8 \left (a \sqrt [3]{x}+b\right )}+\frac {3 a^6}{2 b^7 \left (a \sqrt [3]{x}+b\right )^2}+\frac {63 a^5}{b^8 \sqrt [3]{x}}-\frac {45 a^4}{2 b^7 x^{2/3}}+\frac {10 a^3}{b^6 x}-\frac {9 a^2}{2 b^5 x^{4/3}}+\frac {9 a}{5 b^4 x^{5/3}}-\frac {1}{2 b^3 x^2} \]

[Out]

3/2*a^6/b^7/(b+a*x^(1/3))^2+21*a^6/b^8/(b+a*x^(1/3))-1/2/b^3/x^2+9/5*a/b^4/x^(5/3)-9/2*a^2/b^5/x^(4/3)+10*a^3/
b^6/x-45/2*a^4/b^7/x^(2/3)+63*a^5/b^8/x^(1/3)-84*a^6*ln(b+a*x^(1/3))/b^9+28*a^6*ln(x)/b^9

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Rubi [A]  time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {263, 266, 44} \[ -\frac {45 a^4}{2 b^7 x^{2/3}}-\frac {9 a^2}{2 b^5 x^{4/3}}+\frac {21 a^6}{b^8 \left (a \sqrt [3]{x}+b\right )}+\frac {3 a^6}{2 b^7 \left (a \sqrt [3]{x}+b\right )^2}+\frac {63 a^5}{b^8 \sqrt [3]{x}}+\frac {10 a^3}{b^6 x}-\frac {84 a^6 \log \left (a \sqrt [3]{x}+b\right )}{b^9}+\frac {28 a^6 \log (x)}{b^9}+\frac {9 a}{5 b^4 x^{5/3}}-\frac {1}{2 b^3 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^(1/3))^3*x^4),x]

[Out]

(3*a^6)/(2*b^7*(b + a*x^(1/3))^2) + (21*a^6)/(b^8*(b + a*x^(1/3))) - 1/(2*b^3*x^2) + (9*a)/(5*b^4*x^(5/3)) - (
9*a^2)/(2*b^5*x^(4/3)) + (10*a^3)/(b^6*x) - (45*a^4)/(2*b^7*x^(2/3)) + (63*a^5)/(b^8*x^(1/3)) - (84*a^6*Log[b
+ a*x^(1/3)])/b^9 + (28*a^6*Log[x])/b^9

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^3 x^4} \, dx &=\int \frac {1}{\left (b+a \sqrt [3]{x}\right )^3 x^3} \, dx\\ &=3 \operatorname {Subst}\left (\int \frac {1}{x^7 (b+a x)^3} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\frac {1}{b^3 x^7}-\frac {3 a}{b^4 x^6}+\frac {6 a^2}{b^5 x^5}-\frac {10 a^3}{b^6 x^4}+\frac {15 a^4}{b^7 x^3}-\frac {21 a^5}{b^8 x^2}+\frac {28 a^6}{b^9 x}-\frac {a^7}{b^7 (b+a x)^3}-\frac {7 a^7}{b^8 (b+a x)^2}-\frac {28 a^7}{b^9 (b+a x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {3 a^6}{2 b^7 \left (b+a \sqrt [3]{x}\right )^2}+\frac {21 a^6}{b^8 \left (b+a \sqrt [3]{x}\right )}-\frac {1}{2 b^3 x^2}+\frac {9 a}{5 b^4 x^{5/3}}-\frac {9 a^2}{2 b^5 x^{4/3}}+\frac {10 a^3}{b^6 x}-\frac {45 a^4}{2 b^7 x^{2/3}}+\frac {63 a^5}{b^8 \sqrt [3]{x}}-\frac {84 a^6 \log \left (b+a \sqrt [3]{x}\right )}{b^9}+\frac {28 a^6 \log (x)}{b^9}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 130, normalized size = 0.89 \[ \frac {-840 a^6 \log \left (a \sqrt [3]{x}+b\right )+280 a^6 \log (x)+\frac {b \left (840 a^7 x^{7/3}+1260 a^6 b x^2+280 a^5 b^2 x^{5/3}-70 a^4 b^3 x^{4/3}+28 a^3 b^4 x-14 a^2 b^5 x^{2/3}+8 a b^6 \sqrt [3]{x}-5 b^7\right )}{x^2 \left (a \sqrt [3]{x}+b\right )^2}}{10 b^9} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^(1/3))^3*x^4),x]

[Out]

((b*(-5*b^7 + 8*a*b^6*x^(1/3) - 14*a^2*b^5*x^(2/3) + 28*a^3*b^4*x - 70*a^4*b^3*x^(4/3) + 280*a^5*b^2*x^(5/3) +
1260*a^6*b*x^2 + 840*a^7*x^(7/3)))/((b + a*x^(1/3))^2*x^2) - 840*a^6*Log[b + a*x^(1/3)] + 280*a^6*Log[x])/(10
*b^9)

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fricas [A]  time = 1.01, size = 230, normalized size = 1.58 \[ \frac {280 \, a^{9} b^{3} x^{3} + 420 \, a^{6} b^{6} x^{2} + 90 \, a^{3} b^{9} x - 5 \, b^{12} - 840 \, {\left (a^{12} x^{4} + 2 \, a^{9} b^{3} x^{3} + a^{6} b^{6} x^{2}\right )} \log \left (a x^{\frac {1}{3}} + b\right ) + 840 \, {\left (a^{12} x^{4} + 2 \, a^{9} b^{3} x^{3} + a^{6} b^{6} x^{2}\right )} \log \left (x^{\frac {1}{3}}\right ) + 15 \, {\left (56 \, a^{11} b x^{3} + 98 \, a^{8} b^{4} x^{2} + 36 \, a^{5} b^{7} x - 3 \, a^{2} b^{10}\right )} x^{\frac {2}{3}} - 3 \, {\left (140 \, a^{10} b^{2} x^{3} + 224 \, a^{7} b^{5} x^{2} + 63 \, a^{4} b^{8} x - 6 \, a b^{11}\right )} x^{\frac {1}{3}}}{10 \, {\left (a^{6} b^{9} x^{4} + 2 \, a^{3} b^{12} x^{3} + b^{15} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^(1/3))^3/x^4,x, algorithm="fricas")

[Out]

1/10*(280*a^9*b^3*x^3 + 420*a^6*b^6*x^2 + 90*a^3*b^9*x - 5*b^12 - 840*(a^12*x^4 + 2*a^9*b^3*x^3 + a^6*b^6*x^2)
*log(a*x^(1/3) + b) + 840*(a^12*x^4 + 2*a^9*b^3*x^3 + a^6*b^6*x^2)*log(x^(1/3)) + 15*(56*a^11*b*x^3 + 98*a^8*b
^4*x^2 + 36*a^5*b^7*x - 3*a^2*b^10)*x^(2/3) - 3*(140*a^10*b^2*x^3 + 224*a^7*b^5*x^2 + 63*a^4*b^8*x - 6*a*b^11)
*x^(1/3))/(a^6*b^9*x^4 + 2*a^3*b^12*x^3 + b^15*x^2)

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giac [A]  time = 0.21, size = 123, normalized size = 0.84 \[ -\frac {84 \, a^{6} \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{b^{9}} + \frac {28 \, a^{6} \log \left ({\left | x \right |}\right )}{b^{9}} + \frac {840 \, a^{7} b x^{\frac {7}{3}} + 1260 \, a^{6} b^{2} x^{2} + 280 \, a^{5} b^{3} x^{\frac {5}{3}} - 70 \, a^{4} b^{4} x^{\frac {4}{3}} + 28 \, a^{3} b^{5} x - 14 \, a^{2} b^{6} x^{\frac {2}{3}} + 8 \, a b^{7} x^{\frac {1}{3}} - 5 \, b^{8}}{10 \, {\left (a x^{\frac {1}{3}} + b\right )}^{2} b^{9} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^(1/3))^3/x^4,x, algorithm="giac")

[Out]

-84*a^6*log(abs(a*x^(1/3) + b))/b^9 + 28*a^6*log(abs(x))/b^9 + 1/10*(840*a^7*b*x^(7/3) + 1260*a^6*b^2*x^2 + 28
0*a^5*b^3*x^(5/3) - 70*a^4*b^4*x^(4/3) + 28*a^3*b^5*x - 14*a^2*b^6*x^(2/3) + 8*a*b^7*x^(1/3) - 5*b^8)/((a*x^(1
/3) + b)^2*b^9*x^2)

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maple [A]  time = 0.01, size = 123, normalized size = 0.84 \[ \frac {3 a^{6}}{2 \left (a \,x^{\frac {1}{3}}+b \right )^{2} b^{7}}+\frac {21 a^{6}}{\left (a \,x^{\frac {1}{3}}+b \right ) b^{8}}+\frac {28 a^{6} \ln \relax (x )}{b^{9}}-\frac {84 a^{6} \ln \left (a \,x^{\frac {1}{3}}+b \right )}{b^{9}}+\frac {63 a^{5}}{b^{8} x^{\frac {1}{3}}}-\frac {45 a^{4}}{2 b^{7} x^{\frac {2}{3}}}+\frac {10 a^{3}}{b^{6} x}-\frac {9 a^{2}}{2 b^{5} x^{\frac {4}{3}}}+\frac {9 a}{5 b^{4} x^{\frac {5}{3}}}-\frac {1}{2 b^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^(1/3))^3/x^4,x)

[Out]

3/2*a^6/b^7/(a*x^(1/3)+b)^2+21*a^6/b^8/(a*x^(1/3)+b)-1/2/b^3/x^2+9/5*a/b^4/x^(5/3)-9/2*a^2/b^5/x^(4/3)+10*a^3/
b^6/x-45/2*a^4/b^7/x^(2/3)+63*a^5/b^8/x^(1/3)-84*a^6*ln(a*x^(1/3)+b)/b^9+28*a^6*ln(x)/b^9

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maxima [A]  time = 0.49, size = 146, normalized size = 1.00 \[ -\frac {84 \, a^{6} \log \left (a + \frac {b}{x^{\frac {1}{3}}}\right )}{b^{9}} - \frac {{\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{6}}{2 \, b^{9}} + \frac {24 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{5} a}{5 \, b^{9}} - \frac {21 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{4} a^{2}}{b^{9}} + \frac {56 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{3} a^{3}}{b^{9}} - \frac {105 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{2} a^{4}}{b^{9}} + \frac {168 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} a^{5}}{b^{9}} - \frac {24 \, a^{7}}{{\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} b^{9}} + \frac {3 \, a^{8}}{2 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{2} b^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^(1/3))^3/x^4,x, algorithm="maxima")

[Out]

-84*a^6*log(a + b/x^(1/3))/b^9 - 1/2*(a + b/x^(1/3))^6/b^9 + 24/5*(a + b/x^(1/3))^5*a/b^9 - 21*(a + b/x^(1/3))
^4*a^2/b^9 + 56*(a + b/x^(1/3))^3*a^3/b^9 - 105*(a + b/x^(1/3))^2*a^4/b^9 + 168*(a + b/x^(1/3))*a^5/b^9 - 24*a
^7/((a + b/x^(1/3))*b^9) + 3/2*a^8/((a + b/x^(1/3))^2*b^9)

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mupad [B]  time = 1.21, size = 125, normalized size = 0.86 \[ \frac {\frac {4\,a\,x^{1/3}}{5\,b^2}-\frac {1}{2\,b}+\frac {14\,a^3\,x}{5\,b^4}-\frac {7\,a^2\,x^{2/3}}{5\,b^3}+\frac {126\,a^6\,x^2}{b^7}-\frac {7\,a^4\,x^{4/3}}{b^5}+\frac {28\,a^5\,x^{5/3}}{b^6}+\frac {84\,a^7\,x^{7/3}}{b^8}}{a^2\,x^{8/3}+b^2\,x^2+2\,a\,b\,x^{7/3}}-\frac {168\,a^6\,\mathrm {atanh}\left (\frac {2\,a\,x^{1/3}}{b}+1\right )}{b^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b/x^(1/3))^3),x)

[Out]

((4*a*x^(1/3))/(5*b^2) - 1/(2*b) + (14*a^3*x)/(5*b^4) - (7*a^2*x^(2/3))/(5*b^3) + (126*a^6*x^2)/b^7 - (7*a^4*x
^(4/3))/b^5 + (28*a^5*x^(5/3))/b^6 + (84*a^7*x^(7/3))/b^8)/(a^2*x^(8/3) + b^2*x^2 + 2*a*b*x^(7/3)) - (168*a^6*
atanh((2*a*x^(1/3))/b + 1))/b^9

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sympy [A]  time = 22.23, size = 707, normalized size = 4.84 \[ \begin {cases} \frac {\tilde {\infty }}{x^{2}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {1}{2 b^{3} x^{2}} & \text {for}\: a = 0 \\- \frac {1}{3 a^{3} x^{3}} & \text {for}\: b = 0 \\\frac {280 a^{8} x^{\frac {13}{3}} \log {\relax (x )}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} - \frac {840 a^{8} x^{\frac {13}{3}} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {560 a^{7} b x^{4} \log {\relax (x )}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} - \frac {1680 a^{7} b x^{4} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {840 a^{7} b x^{4}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {280 a^{6} b^{2} x^{\frac {11}{3}} \log {\relax (x )}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} - \frac {840 a^{6} b^{2} x^{\frac {11}{3}} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {1260 a^{6} b^{2} x^{\frac {11}{3}}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {280 a^{5} b^{3} x^{\frac {10}{3}}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} - \frac {70 a^{4} b^{4} x^{3}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {28 a^{3} b^{5} x^{\frac {8}{3}}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} - \frac {14 a^{2} b^{6} x^{\frac {7}{3}}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} + \frac {8 a b^{7} x^{2}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} - \frac {5 b^{8} x^{\frac {5}{3}}}{10 a^{2} b^{9} x^{\frac {13}{3}} + 20 a b^{10} x^{4} + 10 b^{11} x^{\frac {11}{3}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**(1/3))**3/x**4,x)

[Out]

Piecewise((zoo/x**2, Eq(a, 0) & Eq(b, 0)), (-1/(2*b**3*x**2), Eq(a, 0)), (-1/(3*a**3*x**3), Eq(b, 0)), (280*a*
*8*x**(13/3)*log(x)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) - 840*a**8*x**(13/3)*log(x
**(1/3) + b/a)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) + 560*a**7*b*x**4*log(x)/(10*a*
*2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) - 1680*a**7*b*x**4*log(x**(1/3) + b/a)/(10*a**2*b**9
*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) + 840*a**7*b*x**4/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4
+ 10*b**11*x**(11/3)) + 280*a**6*b**2*x**(11/3)*log(x)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x
**(11/3)) - 840*a**6*b**2*x**(11/3)*log(x**(1/3) + b/a)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x
**(11/3)) + 1260*a**6*b**2*x**(11/3)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) + 280*a**
5*b**3*x**(10/3)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) - 70*a**4*b**4*x**3/(10*a**2*
b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) + 28*a**3*b**5*x**(8/3)/(10*a**2*b**9*x**(13/3) + 20*a*
b**10*x**4 + 10*b**11*x**(11/3)) - 14*a**2*b**6*x**(7/3)/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*
x**(11/3)) + 8*a*b**7*x**2/(10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)) - 5*b**8*x**(5/3)/(
10*a**2*b**9*x**(13/3) + 20*a*b**10*x**4 + 10*b**11*x**(11/3)), True))

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